Uniformly Continuous Function to Complete Metric Space has Unique Continuous Extension to Closure of Domain
Theorem
Let $\tuple {X, d}$ be a metric space.
Let $\tuple {Y, d'}$ be a complete metric space.
Let $A \subseteq X$.
Let $f : A \to Y$ be a uniformly continuous function.
Then there exists a unique continuous function $g : A^- \to Y$ such that:
- $\map g a = \map f a$
for all $a \in A$, where $A^-$ denotes the topological closure of $A$.
Furthermore, $g$ is uniformly continuous.
Proof Outline
Suppose we had a uniformly continuous function $g$ satisfying the hypotheses.
We know that from Sequential Continuity is Equivalent to Continuity in Metric Space, that if $\sequence {a_n}$ is a sequence in $A$ converging to $a \in A^-$, we must have:
- $\map g {a_n} \to \map g a$
Since $\sequence {a_n}$ is a sequence in $A$, this is equivalent to:
- $\map f {a_n} \to \map g a$
So our first goal is to "fill in" the missing values by considering limits of sequences of the form $\sequence {\map f {a_n} }$ where $a_n \to a$.
To ensure uniqueness, we show that the limit is the same regardless of the choice of $\sequence {a_n}$.
We show that the thus constructed function is uniformly continuous, and so is our desired function.
Proof
Existence
Note that if $A$ is closed, then from Set is Closed iff Equals Topological Closure, we have:
- $A^- = A$
So taking $g = f$ suffices in this case.
Suppose now that $A$ is not closed.
Lemma 1
Let $\sequence {a_n}$ be a sequence in $A$ convergent to $a \in A^-$.
Then $\sequence {\map f {a_n} }$ converges.
$\Box$
We now show that the limit of $\sequence {\map f {a_n} }$ is independent of the particular $\sequence {a_n}$ chosen.
Lemma 2
Let $\sequence {a_n}$ be a convergent sequence in $A$.
Then the limit of $\sequence {\map f {a_n} }$ is dependent only on the limit of $\sequence {a_n}$.
That is, there exists a function $L : A^- \to Y$ such that:
- $\ds \lim_{n \mathop \to \infty} \map f {a_n} = \map L {\lim_{n \mathop \to \infty} a_n}$
for every convergent sequence $\sequence {a_n}$.
$\Box$
Now, define the function $g : A^- \to Y$ by:
- $\map g a = \begin{cases}\map L a & a \in A^- \setminus A \\ \map f a & \text{otherwise}\end{cases}$
We now show this function is uniformly continuous.
Let $\epsilon > 0$.
We want to show that there exists some $\delta' > 0$ such that whenever $a, b \in A^-$ and:
- $\map d {a, b} < \delta'$
we have:
- $\map {d'} {\map g a, \map g b} < \epsilon$
From Point in Closure of Subset of Metric Space iff Limit of Sequence:
- there exists a sequence $\sequence {a_n}$ in $A$ such that $a_n \to a$
and:
- there exists a sequence $\sequence {b_n}$ in $A$ such that $b_n \to b$.
We then have, by the Triangle Inequality:
- $\map {d'} {\map g a, \map g b} \le \map {d'} {\map g a, \map f {a_n} } + \map {d'} {\map f {a_n}, \map f {b_n} } + \map {d'} {\map f {b_n}, \map g b}$
Recall by construction of $g$:
- $\map f {a_n} \to \map g a$
and:
- $\map f {b_n} \to \map g b$
From the definition of convergence in a metric space, we can find $N_1 \in \N$ such that:
- $\map {d'} {\map g a, \map f {a_n} } < \epsilon/3$
for $n > N_1$.
We can also find $N_2 \in \N$ such that:
- $\map {d'} {\map g b, \map f {b_n} } < \epsilon/3$
for $n > N_2$.
We now wish to find $N_3 \in \N$ such that:
- $\map {d'} {\map f {a_n}, \map f {b_n} } < \epsilon/3$
for $n > N_3$.
Recall that since $f$ is uniformly continuous function, we can find $\delta > 0$ such that whenever:
- $\map d {a_n, b_n} < \delta$
we have:
- $\map {d'} {\map f {a_n}, \map f {b_n} } < \epsilon/3$
Then by the Triangle Inequality we have:
- $\map d {a_n, b_n} \le \map d {a_n, a} + \map d {a, b} + \map d {b, b_n}$
Since $a_n \to a$ we can pick $M_1 \in \N$ such that:
- $\map d {a_n, a} < \delta/3$
for $n > M_1$.
Since $b_n \to b$ we can pick $M_2 \in \N$ such that:
- $\map d{b, b_n} < \delta/3$
for $n > M_2$.
So, if:
- $\map d {a, b} < \delta/3$
and:
- $n > \max \set {M_1, M_2} = N_3$
we have:
- $\map d {a_n, b_n} < \delta$
and so:
- $\map {d'} {\map f {a_n}, \map f {b_n} } < \epsilon/3$
Now, let:
- $N = \max \set {N_1, N_2, N_3}$
Then for $n > N$ and whenever $\map d {a, b} < \delta/3$, we have:
- $\map {d'} {\map g a, \map g b} \le \map {d'} {\map g a, \map f {a_n} } + \map {d'} {\map f {a_n}, \map f {b_n} } + \map {d'} {\map f {b_n}, \map g b} < \epsilon$
So:
- $\map {d'} {\map g a, \map g b} < \epsilon$
whenever:
- $\map d {a, b} < \delta/3$
So $g$ is uniformly continuous.
From Uniformly Continuous Function is Continuous: Metric Space, $g$ is also continuous.
$\Box$
Uniqueness
Let $h$ be another continuous function satisfying the hypotheses.
Then, for any $a \in A^-$ and sequence $\sequence {a_n}$ convergent sequence to $a$, we have:
- $\map h {a_n} \to \map h a$
and:
- $\map f {a_n} \to \map g a$
by Sequential Continuity is Equivalent to Continuity in Metric Space.
But since $h$ coincides with $f$ on $A$, this gives:
- $\map f {a_n} \to \map h a$
from Convergent Sequence in Metric Space has Unique Limit, we obtain:
- $\map h a = \map g a$
So we must have $g = h$.
$\blacksquare$